because it wont make scenes when its already 180
Answer:
Area of the rectangle = 700 - 2w²
Step-by-step explanation:
In the picture attached one side of the rectangle is Barn.
Two perpendicular sides of the barn are width 'w' and one parallel side to the barn is 'l'.
Jamal has the fencing having length = 700 feet
So, to cover three sides of the rectangle length of wire required = (2w + l) feet
This length should be equal to the length of the wire, which is to be used to cover these three sides.
(2w + l) = 700
l = 700 - 2w ------(1)
Area of the rectangle = lw
By replacing the value of l from equation 1
Area = w(700 - 2w)
Area = 700w - 2w²
Answer:
13 over 3
Step-by-step explanation:
Hi Jakeyriabryant! I hope you’re fine!
I hope I have understood the problem well.
If so, what the exercise raises is the following equality:
(x-1) / 5 = 2/3
From this equation you must clear the "x".
First, we pass the 5 that is dividing on the side of the x, to the other side and passes multiplying
(X – 1) / 5 = 2/3
(X – 1) = (2/3)*5
X – 1 = 10/3
Then we pass the one that is subtracting from the side of the x, to the other side and passes adding
X = 10/3 + 1
Remember that to add or subtract fractions they must have the same denominator or a common denominator (in this case we can write 1 as fraction 3/3). Then,
X = 10/3 + 3/3
X = 13/3
I hope I've been helpful!
Regards!
Okay so I think you go south now give me ma points
Answer:
Hello,
I have reply too quick in comments (sorry)
Step-by-step explanation:
![\left\{\begin{array}{ccc}log_x (27)+log_y (4)=5\\log_x (27)-log_y (4)=1\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}2*log_x (27)=6\\2*log_y (4)=4\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}log_x (27)=3\\log_y (4)=2\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}x^{log_x (27)}=x^3\\y^{log_y (4)}=y^2\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}27=x^3\\4=y^2\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}x=3\\y=2\\\end{array}\right.\\\\](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dlog_x%20%2827%29%2Blog_y%20%284%29%3D5%5C%5Clog_x%20%2827%29-log_y%20%284%29%3D1%5C%5C%5Cend%7Barray%7D%5Cright.%5C%5C%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D2%2Alog_x%20%2827%29%3D6%5C%5C2%2Alog_y%20%284%29%3D4%5C%5C%5Cend%7Barray%7D%5Cright.%5C%5C%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dlog_x%20%2827%29%3D3%5C%5Clog_y%20%284%29%3D2%5C%5C%5Cend%7Barray%7D%5Cright.%5C%5C%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%5E%7Blog_x%20%2827%29%7D%3Dx%5E3%5C%5Cy%5E%7Blog_y%20%284%29%7D%3Dy%5E2%5C%5C%5Cend%7Barray%7D%5Cright.%5C%5C%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7D27%3Dx%5E3%5C%5C4%3Dy%5E2%5C%5C%5Cend%7Barray%7D%5Cright.%5C%5C%5C%5C%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bccc%7Dx%3D3%5C%5Cy%3D2%5C%5C%5Cend%7Barray%7D%5Cright.%5C%5C%5C%5C)